Does the coproduct preserve limits? Let $\mathcal{C}$ be a category with coproducts (+) and all limits.
For a diagram $D : \mathcal{I} \to \mathcal{C}$ is the following true?
$A+ \varprojlim_{i \in \mathcal{I}}  D(i) \cong \varprojlim_{i \in \mathcal{I}} (A + D(i))$
In words, is it true that the coproduct preserves limits? If not, would it be true that it preserves $\omega$-limits? The reason why I think it should be is because $LA \cong A + LA$ has a greatest solution, but I can't seem to be able to show that $A+(-)$ is continuous.
 A: You're asking about the functor $F\colon C\to C$ given by $F(X) = X+A$ for a fixed object $A\in C$. When $C = \mathsf{Set}$, this functor does indeed preserve $\omega$-limits, which is sufficient for the construction of a terminal $F$-coalgebra (a greatest fixed-point of $F$).
More generally, whenever $C$ is an extensive category, $F$ preserves connected limits. To see this, note that for any connected diagram with objects $(X_i)_{i\in I}$ in $C$, there is a natural map $(\varprojlim X_i)+A\to \varprojlim(X_i+A)$ coming from the universal properties of the limit and the coproduct. We just need to construct the inverse map $\varprojlim(X_i+A)\to (\varprojlim X_i)+A$. Let $Z =\varprojlim(X_i+A)$. For all $i$, we have the projection map $\pi_i\colon Z\to (X_i+A)$. Since $C$ is extensive, pulling back the inclusions $X_i\to (X_i+A)$ and $A\to (X_i+A)$ along $\pi_i$ gives subobjects $B_i\to Z$ and $C_i\to Z$ such that $Z\cong B_i+C_i$, with arrows $f_i\colon B_i\to X_i$ and $g_i\colon C_i\to A$. Now whenever there is an arrow $h\colon X_i\to X_j$ in the diagram, we get a big diagram in which all the squares are pullbacks:
$$\require{AMScd}
\begin{CD}
B_i @>>> Z @<<< C_i\\
@V{f_i}VV @V{\pi_i}VV @VV{g_i}V\\
X_i @>>> X_i+A @<<< A\\
@V{h}VV @V{h+\text{id}_A}VV @VV{\text{id}_A}V\\
X_j @>>> X_j+A @<<< A
\end{CD}$$
It follows that $B_i = B_j$, $C_i = C_j$, $h\circ f_i = f_j$, and $g_i = g_j$. Since the diagram is connected, all the $B_i$ are equal to a single object $B$, and the arrows $f_i$ give a cone over the diagram $(X_i)_{i\in I}$, which induces an arrow $f\colon B\to \varprojlim X_i$. And since the diagram is connected, all the $C_i$ are equal to a single object $C$, and all the $g_i$ are equal to a single arrow $g\colon C\to A$. The arrows $f$ and $g$ induce an arrow $f+g\colon Z\cong B+C \to (\varprojlim X_i) +A$, as desired.
On the other hand, as shown in the comments, $F$ does not preserve arbitrary limits, even when $C = \mathsf{Set}$. And for an arbitrary category $C$, $F$ does not even preserve $\omega$-limits. For example, let $C$ be a poset with $(X_i)_{i\in \omega}$ a descending chain of elements with meet $\bot$, and $A$ a non-top element such that $A\vee X_i = \top$ for all $i$. Then $A\vee \inf(X_i) = A$, but $\inf(A\vee X_i) = \top$.
A: The functor (A + -) does not preserve all limits, but it does preserve the limit for the diagram $D : \omega^\text{op} \to \mathcal{C}$ with $D(i) = F^{i}_A 1$, $D(f : i \leftarrow i+1) = F^{i}_A f$ and $F_A X = A + X$. For $i=0$, $D(f) = !$.
The trick is to notice that $\lim_{n \in \omega^\text{op}} F^{n}_A 1$ in Set  contains only elements of the form $\langle *, *, \dots, *, a, a, a, \dots \rangle $ where $a \in A$. In particular, the first element $a \neq *$ determines all the following ones. (1 in Set is $\{*\}$). This is because the projection maps of the limit have to commute with the maps in the sequence.
Now, $\Phi : (A + \lim_{n \in \omega^\text{op}} F^{n}_A 1) \to \lim_{n \in \omega^\text{op}} F^{n}_A 1$  is the unique map into the limit noticing that
$A + \lim_{n \in \omega^\text{op}} F^{n}_A 1$ is a cone for the diagram $D$ (by applying $(A + -)$ to the limit and adding the map into $1$).
The other direction $\Psi :  \lim_{n \in \omega^\text{op}} F^{n}_A 1 \to (A + \lim_{n \in \omega^\text{op}} F^{n}_A 1)$ is obtained by
taking an element of the limit
$l = \langle *, *, \dots, *, a, a, a, \dots \rangle$ to $\text{inj}_1 (a)$ if the second component is $a$ or $\text{inj}_2 (l)$ otherwise (where $\mathit{inj}_i$ are the injections into the coproduct).
It is easy to check that $\Phi$ and $\Psi$ are inverses by using the fact that once an element is fixed it cannot be changed.
